The general solution of the differential equa | vedclass

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Question

(B) Given the differential equation: $\frac{y dx - x dy}{y} = 0$. Since $y 
eq 0$,we can multiply by $y$ to get $y dx - x dy = 0$. Rearranging the te

Vedclass · Accepted Answer

$y = Cx$

Vedclass · Answer

(B) Given the differential equation: $\frac{y dx - x dy}{y} = 0$. Since $y 
eq 0$,we can multiply by $y$ to get $y dx - x dy = 0$. Rearranging the terms,we have $y dx = x dy$. Separating the variables,we get $\frac{dx}{x} = \frac{dy}{y}$. Integrating both sides,we obtain $\int \frac{dx}{x} = \int \frac{dy}{y}$. This results in $\ln|x| = \ln|y| + \ln|C|$,where $\ln|C|$ is the constant of integration. Using the property of logarithms,$\ln|x| = \ln|Cy|$. Taking the exponential of both sides,we get $x = Cy$,which can be rewritten as $y = \frac{1}{C} x$. Since $\frac{1}{C}$ is also an arbitrary constant,we can write $y = Cx$.

The general solution of the differential equation $\frac{y dx - x dy}{y} = 0$ is . . . . . . .

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